Mountain Bike Inflection Points

I rode my mountain bike up to Bull Gap last Sunday.  The ride is about ten miles with a  fairly steady climb of around 3000 feet.  After a few miles I started to get impatient.  I wanted to arrive at the point where  my pedaling  would get easier – where the grade would start to get less steep and I could use a higher gear.  I had a fair idea where the physical inflection point was (past the top of Toothpick trail a mile or so) but as I passed it I was still in my lower gears.  It dawned on me that what I was really waiting for was the effective inflection point – the point at which my tired legs would feel less resistance.  My analysis follows.

The key concept is the idea of effective steepness or slope.  Effective slope is composed of the physical slope of the terrain plus the cumulative effect of all of the climb so far – the sum of all the preceding slopes.

Let f(x) be the terrain elevation as a function of horizontal distance traveled.  Start at (0,0) (f(0) = 0) and reach the end of the climb at (x_e,1) (f(x_e) = 1).  Thus one would pedal a distance of x_e units to climb one unit vertically.  Assume f'(0)=0, f'(x) >0 if x\epsilon(0,x_e),  and f'(x_e) =0. We start on the flats (not the case with my excursion),  climb steadily and reach the top of the hill or at least a momentarily  flat resting place.

I will study two cases.  Case I will be where the cumulative effect of the climb is based on the physical (terrain) slope and Case II where  the effective slopes accumulate.

Let e'(x) stand for the effective slope – a measure of how hard it is to pedal.  e'(x) is the derivative of e(x) which is the “effective” hill – how the climb really feels.  In case I, let e'(x) = f'(x)+k\int{f'(t)dt}.  This is the sum of the physical slope and the cumulative effect of the incline on the legs.  Here k\ge 0 is some constant.  Call k the conditioning constant, a measure how in shape the rider is.  The smaller the k the more fit the rider.  By the fundamental theorem of calculus, e'(x) = f'(x)+kf(x).  Note e'(x)\ge0.

To get a feel for what the constant k could be, consider the maximum slope a fresh rider, namely me, could handle.  At the outside this would be a one in five or 20% grade.  So e'(x)\le 0.2.  At x_e, the top, f'(x_e)=0 and f(x_e)=1.  So if e'(x) = f'(x)+kf(x)\le 0.2 for all points on the ride, k\le 0.2.  This analysis considers e'(x) for k restricted to [0,0.2].

To find the effective inflection point x_i, set e''(x)=0.  So, e''(x_i) = f''(x_i)+kf'(x_i)=0 and \frac{-f''(x_i)}{f'(x_i)} = k>0.  Since f'(x)>0, x_i occurs where f''(x)<0 somewhere “above”  the location of the physical inflection point.  It makes sense. When  f''(x) <0 the slope of the terrain in decreasing and  the rider is tiring as he or she pedals.

Now consider an example.  Let f(x) = ax^3+bx^2+cx+d.  Use x_e = 10 and f(x_e) = 1 and start at (0,0).  This “hill” will have an average slope of 1 in 10, a 10% grade much more than the climb to Bull Gap and more than I can sustain for any length of time, but the numbers are round and the coefficients of the cubic model of the hill don’t get too crazy.  Using the initial and final conditions on f(x) and the fact that the initial and final slopes must be zero,  f(x) = \frac{-1}{500}x^3+\frac{3}{100}x^2.  Note that the terrain inflection point is at x = 5e'(x) = f'(x)+kf(x) now becomes \frac{-6}{500}x^2+\frac{6}{100}+k(\frac{-1}{500}x^3+\frac{3}{100}x^2).  We can integrate e'(x) with initial condition e(0)=0 to get the effective hill and set e''(x) to zero to find the inflection point.  Consider these graphs where k incremented from 0 to 0.2.

Effective Hill - Case I

Notice how the hill becomes much “higher” as k increases.  The blue line is the physical terrain.

Effective Slope - Case I

The maximum slope is occurring farther and farther into the ride.

Effective Inflection Point - Case I

It is easy to see that as a rider’s physical condition worsens, increasing k, the point of easier pedaling gets farther and farther up the hill.

For case II, we accumulate the effective slope to add to the terrain’s slope.  We get this integral equation, e'(x) = f'(x)+k\int{e'(t)dt}.  To solve, take the derivative resulting in e''(x) = f''(x)+ke'(x).  Then solve the resulting differential equation for e'(x).

Using the same terrain model as Case I, gives this non-homogenous linear differential equation, e''(x) = \frac{-6}{500}x+\frac{6}{100}+ke'(x). Wolfram Alpha’s  solution is e'(x)=\frac{3}{250k^2}(e^{kx}(5k-1)-(5k-1)+kx).  The small k keeps the exponential under control.  This function can be integrated to get an effective hill and differentiated to help find effective inflection points.  These graphs are the result.

Effective Hill - Case II

When k gets up to 0.2 the hill is starting to look impossible.

Effective Slope - Case II

Doesn’t look like the pedaling will ever get easier for k=0.2.

Effective Inflection Point - Case II

The effective inflection points are farther up the hill than in Case I and the rider needs to be in better condition to negotiate the hill.

This analysis confirms what my body tells me – the worse my physical conditioning  the longer the hill seems.  Thinking math thoughts as I pedal distracts me and keeps me patient (see this post for another example).  Now I need another problem to think about – maybe calculating the length of time it takes to reach the effective inflection point as a function of sprocket size.


About jrh794

I am a sixty-five year old math instructor at Southern Oregon University. I taught at the College of the Siskiyous in Weed California for twenty-six years. Prior to that I worked as a computer programmer, carpenter and in various other jobs. I graduated from Rice University in 1967 and have a MS in Operations Research from Stanford. In the past I have hand-built a stone house and taken long solo bicycle tours. Now I ride my mountain bike and play golf for recreation.
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